What is the brute force approach and complexity for House Robber III?

rob(node) = max(value + grandchildren, children), with no memory with time complexity O(2^N) and space complexity O(H).

What is the optimal approach and complexity for House Robber III?

One post-order pass: each house returns best-if-robbed and best-if-skipped, no map with optimal time complexity O(N) and space complexity O(H).

How do you identify the pattern for House Robber III?

Look for key signals in the prompt: "row of houses arranged as a binary tree" indicates Each house has a parent and up to two children. The links decide which choices conflict.; "cannot rob two directly linked houses" indicates A house and its child conflict. Siblings and grandparent-grandchild pairs do not conflict.; "most money" indicates Maximise a sum over choices. Each house is either robbed or skipped..

House Robber III

Medium·LeetCode #337 Pro OnlyMedium

LeetCode

Problem Statement

A thief is robbing a row of houses that are arranged as a binary tree. The root is the entrance. Every other house has exactly one parent house.

Each house holds some money. The thief cannot rob two houses that are directly linked, meaning a house and one of its children. Two siblings, or a grandparent and a grandchild, are not directly linked, so both can be robbed.

Return the most money the thief can take in one night.

Input

The root of a binary tree. It has between 1 and 10^4 nodes. Each node's value is the money in that house.

Output

The largest total the thief can rob without taking two directly linked houses.

Example 1

Input: root = [3, 2, 3, null, 3, null, 1]

Output: 7

Rob the root (3), the 3 under the left child (3), and the 1 under the right child. That is 3 + 3 + 1 = 7. None of these three houses is a parent of another one.

Example 2

Input: root = [3, 4, 5, 1, 3, null, 1]

Output: 9

Rob the two children of the root: 4 and 5. They are siblings, so they are not directly linked. Total 4 + 5 = 9.

Constraints

• The number of nodes is in the range [1, 10^4].
• 0 ≤ Node.val ≤ 10^4

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“row of houses arranged as a binary tree”

→ Each house has a parent and up to two children. The links decide which choices conflict.

“cannot rob two directly linked houses”

→ A house and its child conflict. Siblings and grandparent-grandchild pairs do not conflict.

“most money”

→ Maximise a sum over choices. Each house is either robbed or skipped.

Approach Complexity Ladder

Brute Force — plain recursion on rob(node)O(2^N)O(H)

rob(node) = max(value + grandchildren, children), with no memory

Better — same recursion with a memo mapO(N)O(N)

Save each node's answer in an unordered_map, so each node is solved once

Optimal — return two values per house (take, skip)O(N)O(H)

One post-order pass: each house returns best-if-robbed and best-if-skipped, no map

Brute Force Intuition

For one house, there are two choices. Rob it: take its value and add the best from its grandchildren, because the children are off-limits. Skip it: add the best from its two children, which are free. Return the larger of the two. The same subtrees are solved again and again.

Recognition rule for this sub-pattern:

Each node returns a summary of its subtree (height, best path). The global answer is updated while combining.

Remember this · Tree DFS

Trigger

Paths, heights, diameters, validating structure, ancestors, anything "for every subtree".

Think

If my left and right children already gave me their answers, how do I build mine?

Key question

What does my recursive function return, and what do I pass down?

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